which would be greater a^b or b^a , given b > a?

Answer 1

In order for these expression to be defined, we must have a>=0, b>=0. If a=0, then, for every b>a we have a^b =0 and b^a =0, so that a^b < b^a.

Now, let's suppose and b are positive. We have a^b <= b^a if, and only if, b ln(a) <= a ln(b).Since a and b are positive, it follows a^b <= b^a <==> ln(a)/a <= ln(b)/b.

Let's consider the function f:(0, oo) -> R given by f(x) = ln(x)/x. Then, f'(x) = (1 - ln(x))/(x^2) , so that f'(x) > 0 for x in (0, e), f'(e) = 0 and f'(x) < 0 for x in(e, oo). This implies f is strictly increasing on (0, e], has a global maximum at x* =e, with f(x*) = 1/e, and is strictly decreasing on [e, oo).

if 0
If 0 e, then nothing can be affirmed. We can have ln(a)/a < ln(b)/b, ln(a)/a = ln(b)/b or ln(a)/a > ln(b)/b, so that we can have a^b < b^a, a^b = b^a or a^b > b^a. Without knowing exactly a and b, the simple fact that 0 e allows no conclusion.

Finally, if e <= a < b, the our analysis of the function f show that ln(a)/a > ln(b)/b and, therefore, we can affirm that a^b > b^a.

Answer 2

Depends.

Take a=2 and b=4
2^4=16 4^2=16 So they're equal

Now a=2 and b=3
2^3=8 and 3^2=9 so the second expression is greater

Now a=2 and b=5
2^5=32 and 5^2=25 so the first is greater.

Answer 3

Well take numbers that go with what is given,

b is more than a so you can put 2>1

1^2 = 1 2^1= 2

so the greater one is b^a

Answer 4

given that b > a

b^ a > a^b
or b^ a < a^ b
or b^a = a^b
eg :

let b = 3 and a = 2

a^ b = 2^ 3 = 8
b^a = 3 ^ 2 = 9

here, b^ a > a^b

let b = 5 , a = 2
a^ b = 2^ 5 = 32
b^a = 5 ^ 2 = 25
here a^b > b^a

let b = 4 and a = 2
a^ b= 2^4 = 16
b^ a = 4^2 = 16

here a^ b = b^ a

Answer 5

The function a^b - b^a has the value of 0 where a = b, and where

a = (-b/Log(b)) (ProductLog(-Log(b)/b))

which means that in the region where b > a, for values of a where

a > (-b/Log(b)) (ProductLog(-Log(b)/b))

the expression a^b - b^a is greater than 0, or a^b > b^a

and for values of a where

a < (-b/Log(b)) (ProductLog(-Log(b)/b))

the expression a^b - b^a is less than 0, or a^b < b^a

The function (-b/Log(b)) (ProductLog(-Log(b)/b)) looks approximately like the function e²/b that passes through the critical point (e, e).

Check the wikipedia article on the ProductLog function, also known as the Lambert W function.

Answer 6

a^b is greater given b>a

Answer 7

It depends completely on the value of a and b.
For example:
2^4=16, 4^2=16
So, a^b=b^a
3^4=81, 4^3=64
So, 3^4>4^3
i.e, a^b>b^a
1^2=1, 2^1=2
i.e, 1^2<2^1
So, a^b

Answer 8

if a=2, b=3
then, a^b = 2^3 = 8
b^a = 3^2 = 9
b^a is greater in this case

if a=2 and b=4
then answer is 16 & 16........Equual

if a=2 and b= 5
then answer is 32 and 25........a^b is greater

Therefore answer is "EITHER"

Answer 9

it depends upon the difference between a and b.
if b is greater than a by a large no...the answer will be different and if b is greater than a by a small no., the answer will be different.

Answer 10

let us take x^(1/x)
let y = x^(1/x)
ln y = ln x/x
x ln y = ln x
differentiate both sides
x/y dy/dx + ln y = 1/x
dy/dx = (1/x - ln y)/(x/y)
= (1/x-1/x ln x)/(x/y) = y/x^2(1-ln x)
so x^(1/x) decreases
if x > e

so for a and b > e

a^(1/a) > b(^1/b) if a > b or a^b > b^ a

for a and b < e
a^(1/a) < b(^1/b) if a > b then or a^b < b^ a

if a < e and b > e then more analysis is required